Difference between revisions of "1967 AHSME Problems/Problem 27"

(Created page with "== Problem == Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in <math>3</math> hours and the other in <math>4</...")
 
 
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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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If the candles both have length <math>\ell</math>, then the candle that burns in <math>3</math> hours has a stub of <math>\ell</math> at <math>0</math> minutes, and a stub of <math>0</math> at <math>180</math> minutes.  Since the candle burns at a constant rate (i.e. linearly), the stub length of this candle <math>t</math> minutes after being lit is <math>f(t) = \frac{\ell}{180}(180 - t)</math>, since <math>f(0) = \ell</math> and <math>f(180) = 0</math>.
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Similarly, for the second candle that burns out in <math>240</math> minutes, <math>g(t) = \frac{\ell}{240}(240 - t)</math>
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Since the first candle burns out faster, if the two candles are lighted simultaneously, it will always have a shorter stub.  The problem asks for when <math>g(t) = 2f(t)</math>.  Solving this equation gives:
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<math>\frac{\ell}{240}(240 - t) = 2\frac{\ell}{180}(180 - t)</math>
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<math>240 - t = \frac{480}{180}(180 - t)</math>
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<math>240 - t = 480 - \frac{480}{180}t</math>
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<math>\frac{8}{3} t - t = 480 - 240</math>
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<math>t = \frac{3}{5} \cdot 240</math>
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<math>t = 144</math>
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So, the second candle will have a stub twice as big as the first candle <math>144</math> minutes after they are both lit.  If we want this to happen at <math>4</math> PM, the candles have to be lit <math>144</math> minutes earlier, or <math>2</math> hours and <math>24</math> minutes earlier.  This is at <math>\text{1:36 PM}</math>, which is option <math>\fbox{C}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 13:27, 20 July 2019

Problem

Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in $3$ hours and the other in $4$ hours. At what time P.M. should the candles be lighted so that, at 4 P.M., one stub is twice the length of the other?

$\textbf{(A) 1:24}\qquad \textbf{(B) 1:28}\qquad \textbf{(C) 1:36}\qquad \textbf{(D) 1:40}\qquad \textbf{(E) 1:48}$

Solution

If the candles both have length $\ell$, then the candle that burns in $3$ hours has a stub of $\ell$ at $0$ minutes, and a stub of $0$ at $180$ minutes. Since the candle burns at a constant rate (i.e. linearly), the stub length of this candle $t$ minutes after being lit is $f(t) = \frac{\ell}{180}(180 - t)$, since $f(0) = \ell$ and $f(180) = 0$.

Similarly, for the second candle that burns out in $240$ minutes, $g(t) = \frac{\ell}{240}(240 - t)$

Since the first candle burns out faster, if the two candles are lighted simultaneously, it will always have a shorter stub. The problem asks for when $g(t) = 2f(t)$. Solving this equation gives:

$\frac{\ell}{240}(240 - t) = 2\frac{\ell}{180}(180 - t)$

$240 - t = \frac{480}{180}(180 - t)$

$240 - t = 480 - \frac{480}{180}t$

$\frac{8}{3} t - t = 480 - 240$

$t = \frac{3}{5} \cdot 240$

$t = 144$

So, the second candle will have a stub twice as big as the first candle $144$ minutes after they are both lit. If we want this to happen at $4$ PM, the candles have to be lit $144$ minutes earlier, or $2$ hours and $24$ minutes earlier. This is at $\text{1:36 PM}$, which is option $\fbox{C}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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